A bag contains $2$ red marbles, $2$ green marbles, and $4$ blue marbles. If we choose a marble, then another marble without putting the first one back in the bag, what is the probability that the first marble will be red and the second will be green? Write your answer as a simplified fraction.
Solution: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a red marble and leaving it out. Event B is picking a green marble. Let's take the events one at at time. What is the probability that the first marble chosen will be red? There are $2$ red marbles, and $8$ total, so the probability we will pick a red marble is $\dfrac{2} {8}$ After we take out the first marble, we don't put it back in, so there are only $7$ marbles left. Since the first marble was red, there are still $2$ green marbles left. So, the probability of picking a green marble after taking out a red marble is $\dfrac{2} {7}$ Therefore, the probability of picking a red marble, then a green marble is $\dfrac{2}{8} \cdot \dfrac{2}{7} = \dfrac{1}{14}$